Eggshells are made primarily of calcium carbonate. The acid in vinegar softens the calcium of the eggshell. Adding a little vinegar to the water when boiling eggs is an old cook’s trick that makes them easier to peel.
A link to this experiment is below in the comment box
Answer:
The correct option is: (D) -2.4 kJ/mol
Explanation:
<u>Chemical reaction involved</u>: 2PG ↔ PEP
Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol
Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)
Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)
Reactant concentration: 2PG = 0.5 mM
Product concentration: PEP = 0.1 mM
Reaction quotient: ![Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2](https://tex.z-dn.net/?f=Q_%7Br%7D%20%3D%5Cfrac%7B%5Cleft%20%5B%20PEP%20%5Cright%20%5D%7D%7B%5Cleft%20%5B%202PG%20%5Cright%20%5D%7D%20%3D%20%5Cfrac%7B0.1%20mM%7D%7B0.5%20mM%7D%20%3D%200.2)
<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>
![\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20kJ%2Fmol%20%2B%20%5B2.303%20%5Ctimes%20%288.314%20%5Ctimes%2010%5E%7B-3%7D%20kJ%2F%28K.mol%29%29%5Ctimes%20%28310.15%20K%29%5D%20log%20%280.2%29)
![\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20%2B%20%5B5.938%5D%20%5Ctimes%20%28-0.699%29%20%3D%201.7%20-%204.15%20%3D%20%28-2.45%20kJ%2Fmol%29)
<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>
Answer:
6.23 x 10^23 molecules
Explanation:
First find the number of moles of BH3 from the information given. We know the amount of grams present and we can find the molar mass which is 13.84.
We know that moles is grams divided by molar mass so we get 14.32/13.84 which is 1.03 moles.
Finally, to figure out the number of molecules, we multiply 1.03 by Avogadro's number which is 6.022x10^23 and we get 6.23x10^23 molecules.
Answer:
It displays the static pressure loss over a length of 100 feet for a specific duct size and airflow.
<h2>
What is a ductulator?</h2>
People who work in the HVAC and construction industries use ductulators to size the ductwork in mechanical systems. It enables you to select the volumetric flow rate after selecting the static pressure you wish to keep in the duct. The ductulator offers you a wide range of rectangular duct diameters that will create this when you enter those criteria. It also displays the required diameter for a circular duct. The actual instrument is constructed of sturdy, thick paper, and it contains a rotating piece that lets you select the two settings I previously indicated.
